Question

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy photon. n = 2 → n = 1 n = 2 → n = 3 n = 5 → n = 6 n = 7 → n = 5

Answer 1

The n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

Step-by-step explanation:

`E_n=-13.6* (Z^2)/(n^2)ev`

Formula used for the radius of the
`n^(th)` orbit will be,

where,

`E_n` = energy of
`n^(th)` orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

`E_1=-13.6* (Z^2)/(1^2) eV=-13.6 eV`

Energy of the second orbit in H atom .

`E_2=-13.6* (Z^2)/((2)^2) eV=-3.40 eV`

Energy of the third orbit in H atom .

`E_3=-13.6* (Z^2)/((9)^2) eV=-1.51 eV`

Energy of the fifth orbit in H atom .

`E_5=-13.6* (Z^2)/((2)^2) eV=-0.544 eV`

Energy of the sixth orbit in H atom .

`E_6=-13.6* (Z^2)/((2)^2) eV=-0.378 eV`

Energy of the seventh orbit in H atom .

`E_7=-13.6* (Z^2)/((2)^2) eV=-278 eV`

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

`E=E_3-E_2`

`E=(-1.51 eV) -(-3.40 eV)=1.89 eV`

Energy absorbed when: n = 5 → n = 6

`E'=E_6-E_5`

`E'=(-0.378 eV)-(-0.544 eV) =0.166 eV`

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.